To solve this problem, we first need to calculate the molarity of the oxalic acid dihydrate solution using the given information.
1. Calculate the number of moles of oxalic acid dihydrate:
Mass of oxalic acid dihydrate = 0.4774 g
Molar mass of oxalic acid dihydrate = 126.06 g/mol
Moles of oxalic acid dihydrate = mass / molar mass
Moles = 0.4774 g / 126.06 g/mol
2. Calculate the molarity of oxalic acid dihydrate solution:
Volume of oxalic acid dihydrate solution = 31.94 mL = 0.03194 L
Molarity (M) = moles / volume
Molarity = (0.4774 g / 126.06 g/mol) / 0.03194 L
Now that we have the molarity of the oxalic acid dihydrate solution, we can use it to determine the grams of sodium hydroxide required to prepare the 2.5 L solution.
3. Calculate the moles of sodium hydroxide:
Balanced equation: 2NaOH(aq) + H2C2O4(aq) → 2H2O(l) + Na2C2O4(aq)
According to the equation, the mole ratio of NaOH to H2C2O4 is 2:1.
Moles of NaOH = (2/1) * Moles of H2C2O4
4. Calculate the grams of sodium hydroxide:
Volume of sodium hydroxide solution = 2.5 L
Molarity of sodium hydroxide solution = molarity calculated for H2C2O4
Moles of NaOH = (Molarity of NaOH) * (Volume of NaOH solution)
Grams of NaOH = moles * molar mass of NaOH
Finally, calculate the grams of sodium hydroxide required for the 2.5 L solution using the molarity of the NaOH solution obtained from the titration.
Please substitute the values into the equations to find the result.